If `I_(1)=int_(e )^(e^(2)) (dx)/(logx)"and "I_(2)=int_(1)^(2)(e^(x))/(x)dx`,then

Options

A. `I_(1)=I_(2)`
B. `2I_(1)=I_(2)`
C. `I_(1)=2I_(2)`
D. none of these

Explanation

To solve the problem, we need to evaluate the two integrals \( I_1 \) and \( I_2 \) and compare them.

### Step-by-Step Solution

1. **Evaluate \( I_1 \)**:
\[ I_1 = \int_{e}^{e^2} \frac{dx}{\log x} \]
We can use the substitution \( t = \log x \). Then, we have:
\[ dt = \frac{1}{x} dx \quad \Rightarrow \quad dx = x \, dt = e^t \, dt \]
The limits change as follows:
- When \( x = e \), \( t = \log e = 1 \)
- When \( x = e^2 \), \( t = \log e^2 = 2 \)

Thus, the integral becomes:
\[ I_1 = \int_{1}^{2} \frac{e^t \, dt}{t} \]

2. **Evaluate \( I_2 \)**:
\[ I_2 = \int_{1}^{2} \frac{e^x}{x} \, dx \]
This integral is already in a suitable form for comparison.

3. **Compare \( I_1 \) and \( I_2 \)**:
From the transformations we made, we see that:
\[ I_1 = \int_{1}^{2} \frac{e^t}{t} \, dt = \int_{1}^{2} \frac{e^x}{x} \, dx = I_2 \]

4. **Conclusion**:
Therefore, we conclude that:
\[ I_1 = I_2 \]

### Final Answer:
The correct option is:
- Option 1: \( I_1 = I_2 \)